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      -------- Original Message --------
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            <th align="RIGHT" nowrap="nowrap" valign="BASELINE">Subject:
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            <td>Re: Lab</td>
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            <th align="RIGHT" nowrap="nowrap" valign="BASELINE">Date: </th>
            <td>Sun, 26 Oct 2014 19:00:24 -0400</td>
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            <th align="RIGHT" nowrap="nowrap" valign="BASELINE">From: </th>
            <td>Andrew J. Pounds <a class="moz-txt-link-rfc2396E" href="mailto:pounds_aj@mercer.edu">&lt;pounds_aj@mercer.edu&gt;</a></td>
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            <th align="RIGHT" nowrap="nowrap" valign="BASELINE">Reply-To:
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            <td><a class="moz-txt-link-abbreviated" href="mailto:pounds_aj@mercer.edu">pounds_aj@mercer.edu</a></td>
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      <div class="moz-cite-prefix">The following reaction takes place in
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        <br>
        HCl  + NaOH  -&gt; NaCl + H2O<br>
        <br>
        The density of the NaCl solution is listed on the report form.<br>
        <br>
        <br>
        On 10/26/14 18:56,  wrote:<br>
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      <blockquote
cite="mid:C40B2F181831EF44A88CD73525827803130CFB7809@MERCERMAIL.MercerU.local"
        type="cite">
        <p dir="ltr">So for the mass of the solution for reaction three
          do I need to know the density of HCl? It isn't listed in the
          lab report form so can I just assume it's 1 g/mL?</p>
        <div class="gmail_quote">On Oct 26, 2014 5:44 PM, "Andrew J.
          Pounds" &lt;<a moz-do-not-send="true"
            href="mailto:pounds_aj@mercer.edu">pounds_aj@mercer.edu</a>&gt;

          wrote:<br type="attribution">
          <blockquote class="gmail_quote" style="margin:0 0 0
            .8ex;border-left:1px #ccc solid;padding-left:1ex">
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              <div>On 10/26/14 15:19,  wrote:<br>
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              <blockquote type="cite">
                <div dir="ltr">Hi Dr. Pounds!
                  <div>I'm trying to do the lab, and we either really
                    messed up in lab or I'm calculating the deltaH
                    values for reactions three and four wrong. I think
                    my mistake might be in calculating the mass of the
                    solution for these reactions. Could you give me any
                    pointers for calculating the mass of the solution
                    for reactions three and four?</div>
                  <div>Thanks! <br clear="all">
                    <div> <br>
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              </blockquote>
              <br>
              <font face="serif">So, in the equation <img
                  style="vertical-align:middle"
                  src="cid:part2.09030405.01050909@mercer.edu" alt="$q =
                  m c_s \Delta T$"> you need to use the mass of the
                solution.  You know the volume and the density of the
                solution is listed on the lab report form.  You can
                therefore use <img style="vertical-align:middle"
                  src="cid:part3.06080009.07080703@mercer.edu" alt="$m =
                  \rho V$"> to compute the mass of the solutions.   The
                specific heats for these solutions can also be found on
                the report form.<br>
                <br>
                Now, when it comes to <img
                  style="vertical-align:middle"
                  src="cid:part4.04090901.09040504@mercer.edu"
                  alt="$\Delta H$">, you have to remember that it is the
                constant volume heat capacity on a per mole basis.   So
                take your <img style="vertical-align:middle"
                  src="cid:part5.03060507.03010804@mercer.edu" alt="$q$">
                that you determined and divide it by the moles of
                substance.<br>
                <br>
                In the case of the dissolution of NaOH you have roughly
                2 grams of NaOH.  Convert this to moles  (2 g / 40
                g/mol) = 0.05.  So your enthalpy of reaction for the
                dissolution of NaOH should be <img
                  style="vertical-align:middle"
                  src="cid:part6.07010100.09090006@mercer.edu" alt="$q /
                  0.05$">.<br>
                <br>
                In the case of the reaction of NaOH (aq) with HCl (aq)
                you have 50 ml of 1 M NaOH (that would also be 0.05
                moles), so divide the heat by that amount to get the
                enthalpy like you did before.<br>
                <br>
                Hopefully that give you enought to get through the rest
                of the lab, but let me know if you have any questions.<br>
                <br>
                <br>
                <br>
                <br>
              </font><br>
              <pre cols="72">-- 
Andrew J. Pounds, Ph.D.  (<a moz-do-not-send="true" href="mailto:pounds_aj@mercer.edu" target="_blank">pounds_aj@mercer.edu</a>)
Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   <a moz-do-not-send="true" href="tel:%28478%29%20301-5627" value="+14783015627" target="_blank">(478) 301-5627</a>
<a moz-do-not-send="true" href="http://faculty.mercer.edu/pounds_aj" target="_blank">http://faculty.mercer.edu/pounds_aj</a>
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      <pre class="moz-signature" cols="72">-- 
Andrew J. Pounds, Ph.D.  (<a moz-do-not-send="true" class="moz-txt-link-abbreviated" href="mailto:pounds_aj@mercer.edu">pounds_aj@mercer.edu</a>)
Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627
<a moz-do-not-send="true" class="moz-txt-link-freetext" href="http://faculty.mercer.edu/pounds_aj">http://faculty.mercer.edu/pounds_aj</a>
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